My friend Alex Ryba uses interesting math questions in the CUNY Math
Challenge. For the 2016 challenge they had the following problem.

Problem. Eve owns two six-sided dice. They are not necessarily fair dice and not necessarily weighted in the same
manner. Eve promises to give Alice and Bob each a fabulous prize if they each roll the same sum with her
dice. Eve wishes to design her two dice to minimize the likelihood that she has to buy two fabulous prizes.
Can she weight them so that the probability for Alice and Bob to get prizes is less than 1/10?

The best outcome for Eve would be if she can weight the dice so that the
sum is uniform. In this case the probability that Alice and Bob get the
prizes is 1/11. Unfortunately for Eve, such a distribution of weight
for the dice is impossible. There are many ways to prove it.

I found a beautiful argument by Hagen von Eitzen on the stack exchange: Let ai (correspondingly bi) be the probabilities that die A (correspondingly B) shows i + 1. It would be very useful later that that i ranges over {0,1,2,3,4,5} for both dice. Let f(z) = ∑ aizi and g(z) = ∑ bizi. Then the desired result is that f(z)g(z) = ∑j=010 zj/11.
The roots of the right side are the non-real roots of unity. Therefore
both f and g have no real roots. So, they must both have even degree.
This implies a5=b5=0 and the coefficient of z10 in their product is also 0, contradiction.

Alex himself has a straightforward argument. The probabilities of 2 and 12 have to be equal to 1/11, therefore, a0b0 = a5b5 = 1/11. Then the probability of a total 7 is at least a0b5 + a0b5. The geometric mean of a0b5 and a0b5
is 1/11 (from above), so their arithmetic mean is at least 1/11 and
their sum is at least 2/11. Therefore, the uniform distribution for sums
is impossible.